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NWeid1
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1. Homework Statement
Give a graphical argument that if f(a)=g(a) and f'(x)>g'(x) for all x>a, then f(x)>g(x) for all x>a. Use the Mean Value Theorem to prove it.
2. Homework Equations
3. The Attempt at a Solution
I have sketched a graphical argument to show that f(x)>g(x). This is what I got:
Let h(x) = f(x) - g(x).
Then h'(x) = f'(x) - g'(x) > 0 for any x > a.
Now apply the MVT on the interval [a, x].
So.. h'(c) = (h(x) - h(a))/(x -a)
And then
[tex]h'(c) =\frac{f(x) - f(a) - g(x) + g(a)}{x - a} > 0[/tex]
[tex]h'(c) =f(x) - f(a) - g(x) + g(a) > 0[/tex]
[tex]h'(c) =f(x) - f(a) > g(x) - g(a)[/tex]
[tex]h'(c) =\frac{f(x) - f(a)}{x-a} > \frac{g(x) - g(a)}{x-a}[/tex]
==> h'(c) = f(x) > g(x)
==> f(x) > f(x) for all x>a
But I feel like this is wrong?!
Give a graphical argument that if f(a)=g(a) and f'(x)>g'(x) for all x>a, then f(x)>g(x) for all x>a. Use the Mean Value Theorem to prove it.
2. Homework Equations
3. The Attempt at a Solution
I have sketched a graphical argument to show that f(x)>g(x). This is what I got:
Let h(x) = f(x) - g(x).
Then h'(x) = f'(x) - g'(x) > 0 for any x > a.
Now apply the MVT on the interval [a, x].
So.. h'(c) = (h(x) - h(a))/(x -a)
And then
[tex]h'(c) =\frac{f(x) - f(a) - g(x) + g(a)}{x - a} > 0[/tex]
[tex]h'(c) =f(x) - f(a) - g(x) + g(a) > 0[/tex]
[tex]h'(c) =f(x) - f(a) > g(x) - g(a)[/tex]
[tex]h'(c) =\frac{f(x) - f(a)}{x-a} > \frac{g(x) - g(a)}{x-a}[/tex]
==> h'(c) = f(x) > g(x)
==> f(x) > f(x) for all x>a
But I feel like this is wrong?!